Question

Know that x+y=5 find the maximum of 2x^2+xy-3y^2 ?

Answer 1

Maximum possible value of `2x^2+xy-3y^2` is `625/8` or `78.125` when `x=35/4=8.75`

Explanation:

As `x+y=5`, `y=5-x`

and `2x^2+xy-3y^2`

= `2x^2+x(5-x)-3(5-x)^2`

= `2x^2+5x-x^2-3(25-10x+x^2)`

= `2x^2+5x-x^2-75+30x-3x^2`

= `-2x^2+35x-75`

= `-2(x^2-2xx35/4x+(35/4)^2-(35/4)^2)-75`

= `-2(x-35/4)^2+2xx(35/4)^2-75`

= `-2(x-35/4)^2+1225/8-75`

= `-2(x-35/4)^2+625/8`

= `625/8-2(x-35/4)^2`

Hence maximum possible value of `2x^2+xy-3y^2` is maximum value of `625/8-2(x-35/4)^2`. Now as `(x-35/4)^2` is always positive, this will be when `x-35/4=0` i.e. `x=35/4`

and value will be `625/8`.